Ex 7.3, 11
In how many ways can the letters of the word PERMUTATIONS be arranged if the
words start with P and end with S
Let first position be P & last position be S (both are fixed)
Since letter are repeating
Hence we use this formula 𝑛!/𝑝1!𝑝2!𝑝3!
Total number of letters = n = 10
& Since, 2T
⇒ p1 = 2
Total arrangements = 10!/2!
= 181440
Ex7.3, 11
In how many ways can the letters of the word PERMUTATIONS be arranged if the
(ii) vowels are all together,
Vowels are a, e, i, o, u
Vowels in word PERMUTATION = (E U A I O)
All vowels comes together
treat as a single object
So our letters become
We arrange them now
Total number of arrangement = 8!/2! × 120
= 2419200
Ex7.3, 11
In how many ways can the letters of the word PERMUTATIONS be arranged if the
(iii) there are always 4 letters between P and S?
Hence,
Number of arrangements when P is before S = 7 × 10!/2!
Similarly,
S can be before P
Number of arrangements when S is before P = 7 × 10!/2!
Total number of arrangements = 7 × 10!/2! + 7 × 10!/2!
= 2 × 7 × 10!/2!
= 2 × 7 × 10!/((2 × 1))
= 7 × 10!
= 25401600